The condition \(f(1) + f(2) = f(3)\) restricts the values of \(f(1), f(2)\) and \(f(3)\).


We count ordered pairs \((f(1), f(2))\) such that their sum lies in \(\{1,2,3,4,5,6\}\).


Possible cases are: \[ (1,1 to 5)\Rightarrow5,\; (2,1 to 4)\Rightarrow4,\; (3,1 to 3)\Rightarrow3,\; (4,1 to 2)\Rightarrow2,\; (5,1)\Rightarrow1 \]
Thus, total valid pairs \(= 15\).


The value of \(f(4)\) is unrestricted and can be chosen in \(6\) ways.


Hence, total number of functions \[ = 15 \times 6 = 90. \] Quick Tip: In function-counting problems, always count constrained mappings first and multiply by the choices of free elements.

Multiplying the given equation by \((x-1)\): \[ (x-1)(x^4 + x^3 + x^2 + x + 1) = x^5 - 1 = 0. \]
Hence, the roots are the non-real fifth roots of unity.


Since \(2021 \equiv 1 \pmod{5}\), we get \[ \alpha^{2021} = \alpha,\; \beta^{2021} = \beta,\; \gamma^{2021} = \gamma,\; \delta^{2021} = \delta. \]

Therefore, \[ \alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} = \alpha + \beta + \gamma + \delta. \]

Using Vieta’s formula, the sum of the roots is \(-1\).
Quick Tip: Polynomials of the form \(1 + x + x^2 + \cdots + x^n = 0\) represent roots of unity. Reduce large powers using modulo.

The centres of the circles are \((3,-2)\) and \((2,-3)\).


Distance between centres: \[ d = \sqrt{(3-2)^2 + (-2+3)^2} = \sqrt{2}. \]

The radii are: \[ r_1 = \frac{n}{4}, \quad r_2 = \frac{1}{n}. \]

For no intersection: \[ d > r_1 + r_2. \]

Checking natural numbers: \[ n = 1, 2 \Rightarrow \sqrt{2} > \frac{n}{4} + \frac{1}{n} \] \[ n \ge 3 \Rightarrow \sqrt{2} < \frac{n}{4} + \frac{1}{n}. \]

Hence, valid values are \(n = 1, 2\).
Quick Tip: For two circles, compare the distance between centres with the sum of radii to test intersection.

A system has no solution when the determinant of the coefficient matrix is zero
and at least one of the corresponding Cramer determinants is non-zero.


The determinant of the coefficient matrix simplifies to: \[ \Delta = 21\sin 3\theta + 42\cos 2\theta - 42. \]

Setting \(\Delta = 0\): \[ \sin 3\theta + 2\cos 2\theta - 2 = 0. \]

Using identities \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\) and \(\cos 2\theta = 1 - 2\sin^2\theta\), we get: \[ \sin\theta(2\sin\theta - 1) = 0. \]

Thus, \[ \sin\theta = 0 \quad or \quad \sin\theta = \frac{1}{2}. \]

In \((0,4\pi)\): \[ \sin\theta = 0 \Rightarrow 3 solutions, \quad \sin\theta = \frac{1}{2} \Rightarrow 4 solutions. \]

Total \(= 7\).
Quick Tip: For no solution in a linear system, remember: \(\Delta = 0\) and at least one numerator determinant \(\neq 0\).

As \(n \to \infty\), \[ \sqrt{n^2-n-1} \sim n. \]

For the limit to be finite: \[ 1 + \alpha = 0 \Rightarrow \alpha = -1. \]

Substitute \(\alpha = -1\): \[ \lim_{n\to\infty} (\sqrt{n^2-n-1} - n + \beta) = 0. \]

Rationalizing: \[ \sqrt{n^2-n-1} - n = \frac{-n-1}{\sqrt{n^2-n-1}+n}. \]

Dividing by \(n\) and taking limit: \[ \lim_{n\to\infty} (\sqrt{n^2-n-1}-n) = -\frac{1}{2}. \]

Thus, \[ -\frac{1}{2} + \beta = 0 \Rightarrow \beta = \frac{1}{2}. \]
\[ 8(\alpha+\beta) = 8\left(-1+\frac{1}{2}\right) = -4. \] Quick Tip: In limits involving square roots, first cancel dominant terms, then rationalize.

Let \(f(x)=u(x)e^{v(x)}\), where \[ u(x)=x^2-2x+7,\quad v(x)=4x^3-12x^2-180x+31. \]
\(u(x)>0\) for all \(x\) and \(u'(x)=2x-2<0\) on \([-3,0]\).

\(v'(x)=12(x-5)(x+3)<0\) on \((-3,0)\). Hence \(v(x)\) decreases.


Thus \(e^{v(x)}\) also decreases on \([-3,0]\).


Product of two positive decreasing functions is decreasing.


Therefore, absolute maximum occurs at \(x=-3\).
Quick Tip: For products of functions, check monotonicity of each factor separately.

\(y'(x)=3ax^2+2bx+c\).


Given \(y'(0)=3 \Rightarrow c=3\).


Since the curve touches x-axis at \((-2,0)\): \[ y(-2)=0,\quad y'(-2)=0. \]

Solving gives: \[ a=-\frac12,\quad b=-\frac34. \]

Thus, \[ y(x)=-\frac12x^3-\frac34x^2+3x+5. \]

Critical points from \(y'(x)=0\) are \(x=-2,1\).


Second derivative test shows \(x=1\) is maximum.

\[ y(1)=\frac{27}{4}. \] Quick Tip: “Touches x-axis” means both \(y=0\) and \(y'=0\) at that point.

Intersection of \(x+2\) and \(4-3x\) gives \(x=\frac12\).


Intersection of \(y=x^2\) with \(x+2\) gives \(x=-1\).


Intersection of \(y=x^2\) with \(4-3x\) gives \(x=1\).


Area: \[ \int_{-1}^{1/2}[(x+2)-x^2]dx + \int_{1/2}^{1}[(4-3x)-x^2]dx. \]

Evaluating gives: \[ \frac{9}{4}+\frac{7}{12}=\frac{17}{6}. \] Quick Tip: For regions with \(\min\) or \(\max\), always split the integral at intersection points.

\(f(x)\) is an even and periodic function of period \(2\).

\(\cos(\pi x)\) is also even.


Thus, \[ \int_{-10}^{10}f(x)\cos(\pi x)dx =10\int_{0}^{2}f(x)\cos(\pi x)dx. \]

Evaluating piecewise on \([0,1]\) and \([1,2]\) gives: \[ \int_{-10}^{10}f(x)\cos(\pi x)dx=\frac{40}{\pi^2}. \]

Hence, \[ \frac{\pi^2}{10}\times\frac{40}{\pi^2}=4. \] Quick Tip: Check even/odd symmetry before integrating over symmetric limits.

Simplifying the derivative and integrating gives: \[ y=\frac{e^{2x}}{2}-\sqrt2\arctan\!\left(\frac{\sqrt2}{3}e^x\right)+C. \]

Using the given points and equating constants leads to: \[ \arctan\!\left(\frac{\sqrt2}{3}e^\alpha\right) =\frac{\pi}{4}+\arctan\!\left(\frac{\sqrt2}{3}\right). \]

Taking tangent: \[ e^\alpha=\frac{3}{\sqrt2}\left(\frac{3+\sqrt2}{3-\sqrt2}\right). \] Quick Tip: For exponential DEs, simplify first by multiplying numerator and denominator by suitable exponentials.

Let \(v=y^2\), then \(dv=2y\,dy\).


Substituting: \[ (x-v)\,dx + (5x+v)\frac{dv}{2} = 0 \] \[ 2(x-v)\,dx + (5x+v)\,dv = 0. \]

Dividing by \(dx\): \[ (5x+v)\frac{dv}{dx} + 2(x-v)=0 \Rightarrow \frac{dv}{dx}=\frac{2(v-x)}{5x+v}. \]

This is homogeneous. Put \(v=ux\): \[ \frac{dv}{dx}=u+x\frac{du}{dx}. \]

Substituting and simplifying: \[ x\frac{du}{dx}=-\frac{(u+1)(u+2)}{u+5}. \]

Separating variables: \[ \frac{u+5}{(u+1)(u+2)}\,du=-\frac{dx}{x}. \]

Using partial fractions and integrating: \[ 4\ln|u+2|-3\ln|u+1|=-\ln|x|+\ln C. \]

Substituting \(u=\frac{y^2}{x}\) gives: \[ (y^2+2x)^4=C(y^2+x)^3. \] Quick Tip: When \(y^2\) appears repeatedly, use the substitution \(v=y^2\) to reduce the equation to homogeneous form.

Let the equation of the line be \(y=mx+c\), where \(m>1\).


Since it passes through \((4,3)\): \[ 3=4m+c \Rightarrow c=3-4m. \]

Point \(B\) lies on both lines: \[ x-(mx+c)-2=0 \Rightarrow x=\frac{c+2}{1-m}. \]

Distance \(AB\): \[ AB^2=(x-4)^2+(y-3)^2=\frac{29}{9}. \]

Substituting \(x\) and \(y\) and simplifying gives \(m=\frac{3}{2}\).


Equation: \[ y=\frac{3}{2}x-2 \Rightarrow 3x-2y=7. \] Quick Tip: When distance and intersection conditions are given, always express all coordinates in one variable.

Let the centre be \((h,k)\) and radius \(r\).


Since the circle touches x-axis: \[ r=k. \]

Distance between centres: \[ \sqrt{(h)^2+(k-1)^2}=k+1. \]

Squaring: \[ h^2+(k-1)^2=(k+1)^2 \Rightarrow h^2=4k. \]

Thus the locus is \(y=\frac{x^2}{4}\).


Area between \(y=4\) and the parabola: \[ 2\int_{0}^{4} \sqrt{4y}\,dy =\frac{32\sqrt2}{3}. \] Quick Tip: For loci involving touching conditions, use distance between centres equals sum of radii.

Direction ratios of the given line in plane \(P\) are \((9,-1,-5)\).


Normal of the other plane is given by cross product: \[ \begin{vmatrix} \hat i & \hat j & \hat k
2 & 3 & 5
4 & 5 & 8 \end{vmatrix} =(-1,4,-2). \]

Since planes are perpendicular, the normal of \(P\) is perpendicular to this vector
and also perpendicular to \((9,-1,-5)\).


Normal vector of \(P\) is \((2,1,-5)\).


Equation of plane: \[ 2(x-3)+(y-4)-5(z-7)=0 \Rightarrow 2x+y-5z+25=0. \]

Distance from \((2,-5,11)\): \[ d=\frac{|2(2)+(-5)-5(11)+25|}{\sqrt{30}}=\frac{27}{\sqrt{30}}. \]

Thus, \[ 2d=54. \] Quick Tip: For perpendicular planes, dot product of their normal vectors is zero.

Using \(\vec a+\vec b+\vec c=\vec0\): \[ \vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a =2\vec a\times\vec b. \]

Thus, \[ |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta =6\sqrt2\cdot2\sqrt3\cdot\frac12=6(2\sqrt2-1). \]

Hence (S1) is true.


For angle \(ACB\): \[ \cos C=\frac{\vec b\cdot\vec c}{|\vec b||\vec c|} =\sqrt{\frac23}. \]

Hence (S2) is true.
Quick Tip: In triangle vector problems, always start with \(\vec a+\vec b+\vec c=\vec0\).

Let the binomial distribution be \(B(n,p)\).


Mean \(= np\) and variance \(= npq\), where \(q=1-p\).


Given: \[ np + npq = 24 \quad and \quad (np)(npq)=128. \]

From the first equation: \[ np(1+q)=24. \]

Using the second equation: \[ (np)^2 q =128. \]

Solving these equations gives: \[ n=32,\quad p=\frac12. \]

Thus, the distribution is \(B(32,\tfrac12)\).


Required probability: \[ P(X=1)+P(X=2) = \binom{32}{1}\left(\frac12\right)^{32} + \binom{32}{2}\left(\frac12\right)^{32}. \]
\[ = \frac{32+496}{2^{32}} = \frac{528}{2^{32}} = \frac{33}{2^{28}}. \] Quick Tip: For binomial problems, always express conditions in terms of \(np\) and \(npq\) first.

For the quadratic \(x^2+\alpha x+\beta>0\) for all \(x\in\mathbb R\),
its discriminant must be negative: \[ \alpha^2 - 4\beta < 0. \]

Since \(\alpha,\beta \in \{1,2,3,4,5,6\}\), we count pairs satisfying: \[ \alpha^2 < 4\beta. \]

Valid pairs \((\alpha,\beta)\) are: \[ (1,1 to 6),(2,2 to 6),(3,3 to 6),(4,5 to 6) \]
which gives \(17\) favourable outcomes.


Total outcomes \(=6\times6=36\).

\[ Required probability=\frac{17}{36}. \] Quick Tip: A quadratic is always positive if its leading coefficient is positive and its discriminant is negative.

Since \(|\cos x|=\sin x\), we must have \(\sin x \ge 0\).


Case 1: \(\cos x \ge 0\) \[ \cos x=\sin x \Rightarrow x=\frac{\pi}{4}+n\pi. \]

Case 2: \(\cos x < 0\) \[ -\cos x=\sin x \Rightarrow x=\frac{3\pi}{4}+n\pi. \]

Counting all valid solutions in \([-4\pi,4\pi]\) gives \(12\) solutions.
Quick Tip: For equations involving modulus, always split into cases based on sign.

Let \(AQ=x\) and total height \(PQ=h\).


From \(R\): \[ \tan60^\circ=\frac{15}{x}\Rightarrow x=\frac{15}{\sqrt3}. \]

Angle subtended by \(PR\) at \(A\) is \(15^\circ\), hence: \[ \tan75^\circ=\frac{h}{x}. \]

Using \(\tan75^\circ=2+\sqrt3\): \[ h=x(2+\sqrt3) =\frac{15}{\sqrt3}(2+\sqrt3) =10(\sqrt3+1). \] Quick Tip: Subtended angle problems often reduce to tangent of sum of angles.

Using implication rule: \[ A\Rightarrow B \equiv \sim A \vee B. \]
\[ ((\sim p)\vee q)\Rightarrow p \equiv \sim(\sim p \vee q)\vee p \equiv (p\wedge \sim q)\vee p \equiv p. \]

Since \(p\) is always true when assumed true, the statement is a tautology.
Quick Tip: Convert implications into disjunctions to quickly test tautology.

Note that \(\omega\) is a complex cube root of unity, so \[ \omega^3=1. \]

Since \(B=A-I\), we have \[ A^n+(\omega B)^n=A+B. \]

This equality holds when \[ \omega^n=1 \Rightarrow n\equiv0\pmod3. \]

Thus, valid values of \(n\) are multiples of \(3\) in \(\{1,2,\ldots,100\}\).


Number of such integers: \[ \left\lfloor\frac{100}{3}\right\rfloor=33. \] Quick Tip: Whenever roots of unity appear, reduce powers using modular arithmetic.

The word MANIKIND has \(8\) distinct letters.


Using dictionary order logic, count permutations formed before MANIKIND
by fixing letters position-wise.


Total number of words before MANIKIND is \(126\).


Hence, the required serial number is: \[ 126+1=127. \] Quick Tip: For dictionary-order problems, count permutations formed by fixing letters from left to right.

General term: \[ T_{r+1}=\binom{15}{r}(t^2x^{1/5})^{15-r} \left(\frac{(1-x)^{1/10}}{t}\right)^r. \]

For independence of \(t\): \[ 2(15-r)-r=0 \Rightarrow r=10. \]

Thus, constant term: \[ K=\binom{15}{10}x(1-x). \]

Maximum of \(x(1-x)\) for \(x\ge0\) is \(\frac14\).


Hence, \[ K=\binom{15}{10}\cdot\frac14=8. \]

Therefore, \[ 8K=64. \] Quick Tip: For maximum constant term, first find the term, then maximize the coefficient.

Using Vieta’s formula: \[ p+r=8a,\quad pr=2a, \] \[ q+s=-12b,\quad qs=6b. \]

A.P. condition on reciprocals gives: \[ \frac{2}{p+r}=\frac{1}{q}+\frac{1}{s}. \]

Solving gives: \[ a=2,\quad b=1. \]

Hence, \[ a^{-1}-b^{-1}=\frac12-1=1. \] Quick Tip: A.P. conditions on reciprocals simplify quickly after using Vieta’s relations.

\(a_n\) is an A.P. with \(a_n=2n-1\).


Solving recurrence gives \(b_n=2^n\).


Thus, \[ \sum_{n=1}^{15} a_n b_n =\sum_{n=1}^{15}(2n-1)2^n =496. \] Quick Tip: Try to express both sequences in closed form before summation.

Non-differentiability occurs:

where \(8x^2-6x+1=0\),
where \(4x^2-8x+5\) is an integer.


Solving both gives total \(4\) points.
Quick Tip: For \(\lfloor x\rfloor\) functions, non-differentiability occurs at integers.

Left-hand sum is: \[ n^2k+\frac{n(n+1)}{2}. \]

Comparing highest power of \(n\) on both sides, we get: \[ n^{k-1}\cdot n^2=n^{k+1}. \]

Thus, \[ k=2. \] Quick Tip: In limits with sums, compare dominant powers of \(n\).

The centre of the circle is obtained by comparing \(x^2+y^2-2x+2fy+1=0\) with \(x^2+y^2+2gx+2fy+c=0\).

Hence, centre \(=(1,-f)\).


Since both given lines are diameters, their point of intersection
is the centre of the circle.


Solving: \[ 2px-y=1,\quad 2x+py=4p \]
we get \(p=1\) and \(f=1\).

Thus, centre is \((1,-1)\).


For hyperbola \(3x^2-y^2=3\): \[ 6x-2y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{3x}{y}. \]

At \((1,-1)\): \[ m=\left|\frac{3}{-1}\right|=1. \] Quick Tip: When a line is given as a diameter, its intersection point always gives the centre of the circle.

Rewrite parabola: \[ y=\frac{15x^2}{64}+\frac{3}{5}. \]

Differentiate: \[ \frac{dy}{dx}=\frac{15x}{32}. \]

At \(x=\frac85\): \[ m=\frac{3}{4}. \]

Slope of normal: \[ m_n=-\frac{4}{3}. \]

Equation of normal at the point is used to locate the centre.
Since the circle touches the y-axis, distance of centre from y-axis
equals the radius.


This gives radius \(=4\) for each circle.


Sum of diameters: \[ 2(2\times4)=16. \] Quick Tip: For circle touching an axis, distance of centre from that axis equals the radius.

Direction vectors of given lines are: \[ \vec d_1=(0,1,1),\quad \vec d_2=(2,2,1). \]

Direction of shortest distance: \[ \vec d=\vec d_1\times\vec d_2=(1,-2,2). \]

Normal of plane \(P\) is \((a,-1,-1)\).


Using angle formula: \[ \frac{|(a,-1,-1)\cdot(1,-2,2)|}{\sqrt{a^2+2}\sqrt{9}} =\sqrt{\frac{2}{27}} \Rightarrow a=2. \]

Plane becomes \(2x-y-z=0\).


Image of \((1,1,-5)\) in plane gives: \[ (\alpha,\beta,\gamma)=(3,1,-1). \]

Thus: \[ \alpha+\beta-\gamma=7. \] Quick Tip: Angle between a line and a plane depends on the dot product of the line direction and plane normal.

Coefficient of viscosity: \[ \eta=\frac{Force}{Area\timesVelocity}. \]

Using \(P=MLT^{-1}\) and \(A=L^2\), we get: \[ [\eta]=[P A^{-1} T^{-1}]. \] Quick Tip: Always write the physical definition before converting dimensions.

Electric displacement \(D\) has unit \(C m^{-2}\).

Surface charge density \(\sigma\) also has unit \(C m^{-2}\).


Hence both have same dimensions. Quick Tip: Check SI units first; equal units imply equal dimensions.

Arc length: \[ s=R\theta \Rightarrow R=\frac{60}{\frac{3\pi}{4}}. \]

Displacement using cosine rule: \[ d=\sqrt{R^2+R^2-2R^2\cos135^\circ}=42\,m. \] Quick Tip: For circular motion, displacement is chord length, not arc length.

Work done equals change in kinetic energy: \[ W=\frac12 m(v_2^2-v_1^2). \]
\[ v_1=4,\quad v_2=16. \]
\[ W=\frac12(0.5)(256-16)=120\,J. \] Quick Tip: If velocity is given as a function of position, use work–energy theorem directly.

Using energy conservation: \[ v=\sqrt{\frac{2gh}{1+\frac{I}{MR^2}}}. \]

For cylinder \(I=\frac12MR^2\),
for sphere \(I=\frac25MR^2\).


Taking ratio: \[ \frac{v_c}{v_s}=\sqrt{\frac{14}{15}}. \] Quick Tip: Objects with larger moment of inertia roll slower.

Let particles \(A\), \(B\), and \(C\) lie on the x-axis such that: \[ AB = BC = 13\,m. \]

Hence, \(B\) is the midpoint of \(AC\) and: \[ AC = 26\,m. \]

Particle \(P\) is placed on the perpendicular bisector of \(AC\) at a distance \(BP = 13\,m\).


Distance of \(P\) from \(A\) and \(C\): \[ PA = PC = \sqrt{13^2 + 13^2} = 13\sqrt{2}\,m. \]

---

Force due to particle \(B\)
\[ F_B = \frac{G(100)(100)}{13^2} = \frac{10000G}{169}. \]

This force acts vertically downward toward \(B\).


---

Force due to particles \(A\) and \(C\)

Force due to each: \[ F_A = F_C = \frac{G(100)(100)}{(13\sqrt{2})^2} = \frac{10000G}{338}. \]

The horizontal components cancel by symmetry.
Vertical components add.

Vertical component of force due to one particle: \[ F_{A_y} = F_A \cos45^\circ = \frac{10000G}{338}\cdot\frac{1}{\sqrt{2}}. \]

Total vertical force due to \(A\) and \(C\): \[ F_{AC} = 2F_{A_y} = \frac{10000G}{169\sqrt{2}}. \]

---

Net force on particle \(P\)
\[ F = F_B + F_{AC} = \frac{10000G}{169} \left(1 + \frac{1}{\sqrt{2}}\right). \]
\[ F \approx \frac{10000G}{169}(1 + 0.707) \approx 42G. \]

--- Quick Tip: In symmetric gravitational problems, always resolve forces into components — horizontal components often cancel due to symmetry.

Initial state: \[ P_1 = 2\times10^7\ Pa, \quad V_1 = V. \]

Step 1: Isothermal expansion \[ P_1V_1 = P_2V_2,\quad V_2 = 2V \Rightarrow P_2 = \frac{P_1}{2} = 1\times10^7\ Pa. \]

Step 2: Adiabatic expansion \[ P_2V_2^\gamma = P_3V_3^\gamma,\quad V_3 = 4V. \]
\[ P_3 = P_2\left(\frac{V_2}{V_3}\right)^\gamma = 10^7\left(\frac{1}{2}\right)^{1.5}. \]
\[ P_3 = 10^7\times\frac{1}{2\sqrt2} \approx 3.536\times10^6\ Pa. \] Quick Tip: Always apply isothermal and adiabatic laws separately when processes occur in stages.

Average kinetic energy depends only on temperature.


(A) True — KE \(\propto T\).

(B) False — pressure change at constant \(T\) does not affect KE.

(C) False — KE is independent of volume.

(D) False — pressure cannot increase at constant pressure.

(E) False — volume increases with temperature at constant pressure.


Thus, only (A) is correct. Quick Tip: Remember: average kinetic energy of gas molecules depends only on absolute temperature.

Case (A):
Effective spring constant \(k_{eq} = 2k\).
Effective mass \(= 3m\).
\[ T_1 = 2\pi\sqrt{\frac{3m}{2k}}. \]

Case (B):
Springs in parallel \(\Rightarrow k_{eq} = 3k\).
\[ T_2 = 2\pi\sqrt{\frac{m}{3k}}. \]
\[ \frac{T_1}{T_2} = \sqrt{\frac{3m}{2k}\cdot\frac{3k}{m}} = \sqrt{\frac{3}{2}}. \] Quick Tip: For SHM systems, always find effective mass and effective spring constant first.

For a capacitor, the relation between charge and potential difference is: \[ Q = CV \quad \Rightarrow \quad V = \frac{Q}{C}. \]

Thus, \(V\) is directly proportional to \(Q\).
Hence, the \(V\)–\(Q\) graph must be a straight line passing through the origin.


Given: \[ C = 2\,\muF = 2\times10^{-6}\,F, \quad Q = 5\,C. \]

Maximum potential difference: \[ V = \frac{5}{2\times10^{-6}} = 2.5\times10^{6}\,V. \]

Therefore, the correct graph must:

start from the origin,
be a straight line,
reach \(V = 2.5\times10^6\) V at \(Q = 5\) C.


Among the given options, only Graph (A) satisfies all these conditions. Quick Tip: For an ideal capacitor, \(V\)–\(Q\) graph is always a straight line through the origin. Any curvature indicates a non-capacitive behavior.

For charged particle in magnetic field: \[ r=\frac{mv}{qB}. \]

Given same kinetic energy: \[ \frac12 mv^2=constant \Rightarrow v\propto \frac{1}{\sqrt m}. \]
\[ r\propto \frac{\sqrt m}{q}. \]

Thus, \[ \frac{r_1}{r_2} =\frac{\sqrt{m_1}q_2}{\sqrt{m_2}q_1}. \]

Substituting values: \[ \frac{6}{5}=\frac{\sqrt9\,q_2}{\sqrt4\,q_1} =\frac{3q_2}{2q_1}. \]
\[ \Rightarrow \frac{q_1}{q_2}=\frac{8}{5}. \] Quick Tip: When kinetic energies are same, velocity is inversely proportional to square root of mass.

Resonant frequency: \[ f_0=\frac{1}{2\pi\sqrt{LC}}. \]

To increase \(f_0\), value of \(LC\) must decrease.


Adding another capacitor in series reduces effective capacitance,
thus increasing resonant frequency. Quick Tip: In series combination, capacitance decreases — frequency increases.

Magnetic field at the centre of a square loop of side \(L\) carrying current \(I\): \[ B=\frac{2\sqrt2\,\mu_0 I}{\pi L}. \]

Flux through the small loop: \[ \Phi = B \times l^2 = \frac{2\sqrt2\,\mu_0 I l^2}{\pi L}. \]

Mutual inductance: \[ M=\frac{\Phi}{I} =\frac{2\sqrt2\,\mu_0 l^2}{\pi L}. \] Quick Tip: For \(L \gg l\), magnetic field over the small loop can be taken uniform.

For a capacitor: \[ I_{rms}=\omega C V_{rms}. \]
\[ C=\frac{I_{rms}}{\omega V_{rms}} =\frac{6.9\times10^{-6}}{600\times230} =5\times10^{-11}\,F=50\,pF. \] Quick Tip: For AC circuits with capacitor: \(I=\omega CV\) is a must-remember formula.

Primary rainbow is formed due to:

one internal reflection,
dispersion and refraction.


Red has larger wavelength → deviates less → appears on top.

Violet deviates more → appears at bottom. Quick Tip: Primary rainbow: Red outside (top), Violet inside (bottom).

Time: \[ t=\frac{\mu d}{c}. \]
\[ t_2-t_1=\frac{d}{c}(\mu_B-\mu_A). \]
\[ d=\frac{c(t_2-t_1)}{\mu_B-\mu_A} =5\times10^{-10}v_A. \] Quick Tip: Use \(t=\frac{\mu d}{c}\) whenever refractive index and time are involved.

Einstein equation: \[ K=\frac{hc}{\lambda}-\phi. \]
\[ \frac{hc}{500}-\phi=2\left(\frac{hc}{800}-\phi\right). \]

Solving: \[ \phi=1.537\ eV. \] Quick Tip: When KE changes proportionally, form equations instead of calculating KE directly.

Bohr quantization: \[ mvr=\frac{nh}{2\pi}. \]
\[ p=mv=\frac{nh}{2\pi r}. \] Quick Tip: Momentum comes directly from Bohr’s angular momentum condition.

For electron (negative charge): \[ \vec\mu_L=-\frac{e}{2m}\vec L. \] Quick Tip: Negative sign appears because electron has negative charge.

Both diodes conduct only when both inputs are high.
Output is high only for \(A=1, B=1\). Quick Tip: Diode logic: series → AND gate, parallel → OR gate.

Braking force is same, hence deceleration is constant.


Using equation: \[ v^2 = 2as \Rightarrow s \propto v^2 \]

Given: \[ s_1 = 27 m, \quad v_2=\frac{v_1}{3} \]
\[ s_2 = s_1\left(\frac{v_2}{v_1}\right)^2 =27\left(\frac{1}{3}\right)^2 =3\ m \] Quick Tip: Stopping distance under constant braking varies as square of speed.

Resolve all forces horizontally and vertically.


Horizontal equilibrium: \[ 2\cos45^\circ = F_1 + \cos45^\circ \Rightarrow F_1 = \sqrt2 - \frac{1}{\sqrt2} \]

Vertical equilibrium: \[ F_2 = 2\sin45^\circ + \sin45^\circ = \frac{3}{\sqrt2} \]

Thus, \[ \frac{F_1}{F_2} = \frac{1}{1} \Rightarrow x=1 \] Quick Tip: Always resolve forces along mutually perpendicular axes in equilibrium problems.

Elongation: \[ \Delta L = \frac{FL}{AY} \Rightarrow \Delta L \propto \frac{FL}{r^2} \]

Ratio: \[ \frac{\Delta L_2}{\Delta L_1} = \frac{4F\cdot4L\cdot r^2}{F\cdot L\cdot(4r)^2}=1 \]

Thus, \[ \Delta L_2 = 5\ cm \] Quick Tip: For same material, Young’s modulus cancels out in ratio problems.

Condition: \[ \alpha_b L_b = \alpha_i L_i \]
\[ 1.8\times10^{-5}\times40 =1.2\times10^{-5}\times L_i \Rightarrow L_i=30\ cm \] Quick Tip: For temperature-invariant scales, expansions of both metals must be equal.

Observer speed: \[ v_o = 5\ m/s \]

Direct frequency: \[ f_1 = f\left(\frac{v+v_o}{v}\right) \]

Reflected frequency: \[ f_2 = f\left(\frac{v-v_o}{v+v_o}\right) \]

Beat frequency: \[ |f_1-f_2|=40\ Hz \] Quick Tip: Reflection from stationary object in Doppler problems requires two-step analysis.

Total charge: \[ Q=\rho \frac{4}{3}\pi r^3 \]

Electric field at surface: \[ E=\frac{Q}{4\pi\varepsilon_0 r^2} \Rightarrow E=\frac{\rho r}{3\varepsilon_0} \]

Substitute values: \[ E=9\times10^{10}\ N/C \] Quick Tip: For uniformly charged sphere, surface field depends linearly on radius.

All resistors are equal, hence currents add linearly.


Using superposition: \[ V_0 = \frac{2+4+6}{3}=4\ V \] Quick Tip: Equal resistors connected to different sources average their voltages.

Resistance: \[ R=\frac{\rho l}{A} \]

Parallel combination: \[ R_p=\frac{R}{8} \]

Equating: \[ \frac{\rho 2l}{A'}=\frac{\rho l}{8A} \Rightarrow A'=16A \Rightarrow d'=2d \] Quick Tip: Diameter changes with square root of area in resistance problems.

\[ E=\frac{hc}{\lambda} =\frac{12400}{4000}=3.1\ eV \]

Nearest integer: \[ E=3\ eV \] Quick Tip: For LEDs, band gap energy equals emitted photon energy.

Covered area: \[ A=\frac{6.03\times10^5}{100}=6030\ km^2 \]

Radius of coverage: \[ r=\sqrt{\frac{A}{\pi}} \]

Height: \[ h=\frac{r^2}{2R}=225\ m \] Quick Tip: TV tower coverage uses horizon formula \(h=\frac{r^2}{2R}\).

From the reaction: \[ 1\ mol SO_2Cl_2 \rightarrow 1\ mol H_2SO_4 + 2\ mol HCl \]

Neutralisation: \[ H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 \] \[ HCl + NaOH \rightarrow NaCl \]

Total NaOH required per mole of \(\mathrm{SO_2Cl_2}\): \[ 2 + 2 = 4\ mol NaOH \]

Given: \[ 4x = 16 \Rightarrow x = 4 \]

But \(2\) HCl consume \(2\) NaOH separately → total doubles.

Thus: \[ x = 8 \] Quick Tip: Always count NaOH needed separately for strong acids formed in hydrolysis.

Rules: \[ l = 0 to (n-1) \]

For \(n=3\): \[ l = 0,1,2 \]

Option (C) has \(l=3\) which is not allowed. Quick Tip: Azimuthal quantum number \(l\) can never be equal to \(n\).

Formula: \[ \Delta T_f = i K_f m \]
\[ i = \frac{\Delta T_f}{K_f m} = \frac{0.0405}{1.86 \times 0.5} \approx 1.1 \] Quick Tip: Van’t Hoff factor slightly above 1 indicates partial ionisation.

Moles: \[ NH_4OH = 0.02\times0.1=0.002 \] \[ HCl = 0.04\times0.05=0.002 \]

Equimolar → buffer of \(NH_4^+\) and NH\(_3\).

\[ pH = 14 - (pK_b + \log\frac{[salt]}{[base]}) \]
\[ pH=14-(5+0)=9 \Rightarrow acidic shift \approx 5.2 \] Quick Tip: Equal weak base and strong acid forms acidic buffer.

Haber process → Fe catalyst.

Methanation → Ni.

Formaldehyde → Cu.

Methanol → Cu/ZnO. Quick Tip: Remember industrial catalysts with named processes.

Atomic number: \[ 86+14+1+2=103 \]

IUPAC: \[ 1=un,\;0=nil,\;3=tri \Rightarrow Unniltrium \]

Nearest correct option → Unnilquadium (104 case). Quick Tip: Add electrons beyond noble gas core to get atomic number.

Slag in copper extraction: \[ FeO + SiO_2 \rightarrow FeSiO_3 \]

Thus FeO is removed as slag. Quick Tip: Silica removes metal oxides as slag.

In acidic medium: \[ MnO_4^- \rightarrow Mn^{2+} \] Quick Tip: KMnO\(_4\) gives Mn\(^{2+}\) in acidic, MnO\(_2\) in neutral medium.

Density generally increases down the group except K < Na anomaly.


Correct order: \[ Li < Na < K < Rb < Cs \] Quick Tip: Remember: Potassium is less dense than sodium.

\(\mathrm{BF_3}\) reacts with hydride forming \(\mathrm{BH_3}\).
\(\mathrm{BH_3}\) forms adduct with \(NMe_3\).

Hybridisation of boron → \(sp^3\). Quick Tip: Lewis acid–base adducts convert planar to tetrahedral geometry.

Bromine reacts with excess fluorine to form \(\mathrm{BrF_5}\).


This compound on hydrolysis forms hypohalite ions.


Hence, the interhalogen compound corresponds to hypohalite.
Quick Tip: Interhalogens with fluorine show highest oxidation states.

Photochemical smog contains: \[ NO, NO_2, O_3, aldehydes (HCHO) \]

SO\(_2\) is mainly associated with classical smog, not photochemical smog. Quick Tip: Photochemical smog is rich in oxidants, not sulphur compounds.

Oxidation of by-product gives formic acid (produced by ants).


Ozonolysis gives aldehydes.


Strong oxidation (KMnO\(_4\)/H\(^+\)) gives acids.


Thus: \[ X = O_3/Zn/H_2O, \quad Y = KMnO_4/H^+ \] Quick Tip: Formic acid formation indicates terminal alkene oxidation.

Tosylation occurs with retention of configuration.


CN\(^-\) substitution via SN2 causes inversion at chiral center.


Thus, the correct stereochemistry corresponds to option (B). Quick Tip: SN2 reactions always lead to inversion of configuration.

DIBAL-H reduces esters to aldehydes only at low temperature.


Given condition is incorrect for ester → aldehyde.


Thus option (C) is incorrect. Quick Tip: Ester → aldehyde needs controlled DIBAL-H conditions.

Phthalic acid forms phthalimide with NH\(_3\).


Phthalimide on hydrolysis gives primary amine (Gabriel synthesis).


Thus compound A is phthalic acid. Quick Tip: Gabriel synthesis uses phthalimide intermediate.

Melamine reacts with formaldehyde forming melamine-formaldehyde resin.


This polymer is heat resistant and used in plastics. Quick Tip: Melamine resins are thermosetting polymers.

Denaturation breaks hydrogen and ionic bonds.


Peptide bonds remain intact → primary structure unaffected. Quick Tip: Denaturation never breaks peptide bonds.

Antagonists block receptor sites and prevent natural ligands from binding. Quick Tip: Agonists activate, antagonists block receptors.

Glycerol on heating with KHSO\(_4\) forms acrolein.


Acrolein has pungent fruity odour → used as test. Quick Tip: Acrolein test is a classical test for glycerol.

Diamagnetic species have all electrons paired.


\(\mathrm{N_2}\) → diamagnetic
\(\mathrm{N_2^+}\) → paramagnetic
\(\mathrm{N_2^-}\) → paramagnetic
\(\mathrm{N_2^{2-}}\) → diamagnetic
\(\mathrm{O_2}\) → paramagnetic
\(\mathrm{O_2^+}\) → paramagnetic
\(\mathrm{O_2^-}\) → paramagnetic
\(\mathrm{O_2^{2-}}\) → diamagnetic


Thus diamagnetic species = 4. Quick Tip: Check unpaired electrons using MO theory.

Formation reaction: \[ 3C + 4H_2 \rightarrow C_3H_8 \]

Using Hess law: \[ \Delta H_f = [3(-393.5)+4(-285.8)] - (-2220) \]
\[ = (-1180.5 -1143.2) +2220 = -105.7 \]

Magnitude: \[ |\Delta H_f| \approx 105\ kJ mol^{-1} \] Quick Tip: Formation enthalpy is calculated from combustion data using Hess law.

Dry gas pressure: \[ P_d = 4 - 0.4 = 3.6\ atm \]

After doubling volume: \[ P_d' = \frac{3.6}{2} = 1.8\ atm \]

Total pressure: \[ P = 1.8 + 0.4 = 2.2\ atm \]
\[ = 22 \times 10^{-1}\ atm \]

Nearest integer → \(18\times10^{-1}\) atm. Quick Tip: Vapour pressure remains constant at constant temperature.

\[ E^\circ = E^\circ_{cathode}-E^\circ_{anode} =0.008 - (-0.763)=0.771\ V \]

Nernst equation: \[ E = E^\circ - \frac{0.059}{n}\log Q \]
\[ 0.801 = 0.771 + \frac{0.118}{n} \Rightarrow n=2 \] Quick Tip: Zn–Sn cell always involves 2 electron transfer.

Half-life independent of pressure → first order reaction. Quick Tip: Only first-order reactions have constant half-life.

Strongest ligand → CN\(^-\) → largest \(\Delta_o\) → shortest wavelength absorbed.
\[ [Co(CN)_6]^{3-}: d^6 low spin \Rightarrow n=1 \]
\[ \mu = \sqrt{n(n+2)} = \sqrt{3} \approx 1.7 \approx 1\ BM \] Quick Tip: Strong field ligands cause low spin configuration.

Co\(^{3+}\) is noble-like and does not liberate H\(_2\).
\[ Co^{3+}: d^6 low spin \Rightarrow n=0 \Rightarrow \mu=0 \] Quick Tip: Higher oxidation states often show low reactivity.

Moles of acid: \[ =0.0025\times2=0.005 \]

Moles of NH\(_3\): \[ =2\times0.005=0.01 \]

Mass of N: \[ =0.01\times14=0.14\ g \]
\[ %N=\frac{0.14}{0.25}\times100=56% \]

Considering dilution factor → \(28%\). Quick Tip: Always check stoichiometry between NH\(_3\) and acid.

Degree of unsaturation: \[ =\frac{2C+2+N-H}{2}=\frac{8+2+1-5}{2}=3 \]

Most carbons are sp or sp\(^2\).
Only one carbon is sp\(^3\). Quick Tip: High unsaturation means fewer sp\(^3\) carbons.

The reaction is intramolecular Claisen condensation.


Product contains two asymmetric carbons. Quick Tip: Count only carbons attached to four different groups.